Question

There are two thin spheres A and B of the same material and same thickness. They emit like black bodies. Radius of A is double that of B. A and B of same temperature T. When A and B are kept in a room of temperature T₀ (< T), the ratio of their rates of cooling (rate of fall of temperature) is (assume negligible heat exchange between A and B) –

• A. 2 : 1
• B. 1 : 1
• C. 4 : 1
• D. 8 : 1

Answer 1

Answer: (B)

1 : 1

Answer 2

The rate of heat loss by a thin hollow sphere of thickness 'Dx', mean radius 'r' and made of density 'r' is given by

mS $\frac{dT}{dt}$ = – Ī s A (T⁴ – T₀⁴)

(r 4pp² Dx) S $\frac{dT}{dt}$ = – Ī s 4 p r² (T⁴ – T₀⁴)

Ž $\frac{dT}{dt}$= –$\frac{\in \sigma(T^{4} - T_{0}^{4})}{S\Delta x}$ is independent of radius.

Hence rate of cooling is same for both spheres.